Let $M$ be a smooth $m$-manifold and $N_1$, $N_2$ smooth embedded $k$-submanifolds such that

  • $N_1\cap N_2=\partial N_1=\partial N_2$,
  • for each $x\in N_1\cap N_2$, $T_x N_1=T_x N_2$, and
  • for $x\in N_1\cap N_2$, each vector in $T_x N_1$ is inwards-pointing wrt. $N_1$ iff it is outwards pointing wrt. $N_2$ and vice versa.

Is it true that $N_1\cup N_2$ is a smooth embedded $k$-submanifold with $N_1\cap N_2$ in its interior?

Edit: It follows from mollyerin's answer that $N$ is not necessarily the image of a $C^\infty$ embedding. But I have further bounty questions:

  • Is it at least the image of a $C^1$ embedding?
  • Are there some further simple conditions that I should include so that $N$ is a smooth ($C^\infty$) embedded submanifold?
  • Can I "change" $N$ in some arbitrary small neighborhood of $N_1\cap N_2$ to make the resulting manifold smooth?
  • $\begingroup$ Could you please comment on the downvote? $\endgroup$ Jan 23, 2015 at 21:00
  • $\begingroup$ I didn't downvote, but there is a faction on this site who think it's not okay to ask a question unless you show some work that you've done. $\endgroup$ Jan 23, 2015 at 21:05
  • $\begingroup$ Ok, right, thanks. I will work on the quesiton. $\endgroup$ Jan 23, 2015 at 21:05
  • $\begingroup$ Consider something like this: N1 be the upper branch of $y=x^3$, and N2 be the lower branch of $y=-x^3$. $\endgroup$
    – Xipan Xiao
    Jan 23, 2015 at 23:48
  • $\begingroup$ @XipanXiao Right, thanks. Of course I should add that invards pointing vectors in $T_x N_1$ wrt. $N_1$ are outwards pointing wrt. $N_2$. Now is it sufficient? $\endgroup$ Jan 24, 2015 at 0:03

1 Answer 1


This is false; you'll need assumptions (I'm not sure what good ones are) that ensure that the manifolds glue together "smoothly". As a counterexample, let $f_1: (-\infty, 0] \to \mathbb{R}^2$ be the map $f_1(x) = -x^2$, and let $f_2 : [0, \infty) \to \mathbb{R}^2$ be given by $f_2(x) = x^2$. We of course let $N_1$ be the image of $f_1$ and $N_2$ the image of $f_2$. Then $N_1$ and $N_2$ satisfy your conditions. However, $N = N_1 \cup N_2$ is not a smooth embedded manifold.

[Proof: If it were, there would be a smooth embedding $g : \mathbb{R} \to \mathbb{R}^2$ whose image is $N$. Say $g = (g_1, g_2)$. The map $g_1$ satisfies $g_1'(t) \neq 0$ for all $t$, and so (since it is also smooth), it is a diffeomorphism $\mathbb{R} \to \mathbb{R}$ (by the inverse function theorem). Therefore the map $g \circ g_1^{-1}$ is a smooth embedding $\mathbb{R} \to \mathbb{R}^2$ whose image is $N$, and this map has the form $x \mapsto (x, g_2 \circ g_1^{-1}(x)$), so evidently $g_2 \circ g_1(x) = sign(x) x^2$. But this map is not smooth. ]

  • $\begingroup$ Thanks. Can I at least say that $N$ is a $C^1$-submanifold, or something like that? $\endgroup$ Jan 24, 2015 at 11:12
  • $\begingroup$ Seems like $N$ should be the image of a $C^1$ embedding of a smooth manifold, but the proof isn't immediately obvious to me (I get confused about these things); you might try unaccepting my answer and editing the original question to see if anyone has ideas or a reference. (I will also try thinking about it.) $\endgroup$
    – mollyerin
    Jan 24, 2015 at 11:30
  • $\begingroup$ Thanks. I will leave the acceptance now and offer a bounty tomorrow. $\endgroup$ Jan 24, 2015 at 12:52
  • $\begingroup$ I removed the acceptance, at least temporarily, because I'm afraid that otherwise I can't attract attention.. $\endgroup$ Jan 30, 2015 at 10:49
  • $\begingroup$ @PeterFranek That seems wise :D $\endgroup$
    – mollyerin
    Jan 30, 2015 at 11:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.